Removing place bets after a set number of rolls
December 23rd, 2015 at 2:20:09 PM
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Hello fellow forum folk,
It's my first post to the forum so be nice. (Love the site btw)
I need some help with understanding how/if the house edge changes when you remove place bets before they lose due to a 7-out.
On the WoO site, the Wizard explains how the "The mean number of rolls per shooter is 8.525510."
That made me wonder, is it possible using this knowledge to reduce the house edge by taking place bets down after say 5 rolls(or however many # of rolls less than the 8.5 average) before the shooter 7-outs and the place bet loses? Intuitively, it would seem that over the long run, you would win more bets than lose, but does not letting the bet play out negate any perceived advantage?
I'd love to hear everyone's reasoning/theories/thoughts on the matter.
It's my first post to the forum so be nice. (Love the site btw)
I need some help with understanding how/if the house edge changes when you remove place bets before they lose due to a 7-out.
On the WoO site, the Wizard explains how the "The mean number of rolls per shooter is 8.525510."
That made me wonder, is it possible using this knowledge to reduce the house edge by taking place bets down after say 5 rolls(or however many # of rolls less than the 8.5 average) before the shooter 7-outs and the place bet loses? Intuitively, it would seem that over the long run, you would win more bets than lose, but does not letting the bet play out negate any perceived advantage?
I'd love to hear everyone's reasoning/theories/thoughts on the matter.
December 23rd, 2015 at 2:31:12 PM
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The house edge drops to zero when you remove your bets. (Sorry, just had to say the obvious.)
I see this strategy used all the time: after a shooter rolls five numbers, players remove their bets thinking the 7 is due.
I call it the "reverse five count."
I see this strategy used all the time: after a shooter rolls five numbers, players remove their bets thinking the 7 is due.
I call it the "reverse five count."
December 23rd, 2015 at 7:23:08 PM
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You'll lose less if you don't always have a bet working but the edge doesn’t change. The house edge of the place 6 bet is 1.52% of your bet, regardless of how much that is. Even if it's zero.
If you take a bet off after 5 rolls, even if it hasn't resolved, what you're effectively doing is decreasing the average wager from one unit to less than one. You can figure out what that is, and thus the decrease in expected loss - basically you're sitting out X% of the time and therefore playing X% less.
If you take a bet off after 5 rolls, even if it hasn't resolved, what you're effectively doing is decreasing the average wager from one unit to less than one. You can figure out what that is, and thus the decrease in expected loss - basically you're sitting out X% of the time and therefore playing X% less.
"In my own case, when it seemed to me after a long illness that death was close at hand, I found no little solace in playing constantly at dice."
-- Girolamo Cardano, 1563
December 23rd, 2015 at 8:25:33 PM
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The estimated number of rolls changes as you continue shooting. So, if you've shot five times, the estimated number of rolls increases from the 8.whatever to something more (13?).
This is why I don't like talking about rolls before a seven out. You have a 1/6 chance of sevening out when the lammer is ON. Period. If you want to take bets off to lower variance, fine, but know that that's all you're doing and don't try to explain it with "x amount of rolls left".
This is why I don't like talking about rolls before a seven out. You have a 1/6 chance of sevening out when the lammer is ON. Period. If you want to take bets off to lower variance, fine, but know that that's all you're doing and don't try to explain it with "x amount of rolls left".
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